jmc

Given $k,$ suppose $|k-\sqrt{p}|<\frac{1}{2}.$ Then $$k-\frac{1}{2}<\sqrt{p}<k+\frac{1}{2}.$$Squaring both sides, we get $$k^2-k+\frac{1}{4}<p<k^2+k+\frac{1}{4}.$$Thus, given $k,$ the positive integers $p$ such that $b(p)=k$ are $k^2-k+1,$ $k^2-k+2,$ $\dots ,$ $k^2+k,$ for a total of $2k$ numbers. So, these $2k$ numbers contribute $2k\cdot k=2k^2$ to the sum.

Now, $b(2007)=45,$ so $$\begin{array}{rl}S& =\sum _{p=1}^{2007}b(p)\\ & =\sum _{k=1}^{44}2k^2+\sum _{p=1981}^{2007}45\\ & =2\sum _{k=1}^{44}{k}^2+27\cdot 45\\ & =2\cdot \frac{44\cdot 45\cdot 89}{6}+27\cdot 45\\ & =\boxed{59955}.\end{array}$$